\(\int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx\) [195]

Optimal result

Integrand size = 27, antiderivative size = 143 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \]

[Out]

-8/5*e*(-e*x+d)/(-e^2*x^2+d^2)^(5/2)-4/15*e*(-8*e*x+5*d)/d^2/(-e^2*x^2+d^2)^(3/2)+4*e*arctanh((-e^2*x^2+d^2)^(
1/2)/d)/d^4-1/15*e*(-79*e*x+60*d)/d^4/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^4/x

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {866, 1819, 821, 272, 65, 214} \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=\frac {4 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x} \]

[In]

Int[Sqrt[d^2 - e^2*x^2]/(x^2*(d + e*x)^4),x]

[Out]

(-8*e*(d - e*x))/(5*(d^2 - e^2*x^2)^(5/2)) - (4*e*(5*d - 8*e*x))/(15*d^2*(d^2 - e^2*x^2)^(3/2)) - (e*(60*d - 7
9*e*x))/(15*d^4*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^4*x) + (4*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^4

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d-e x)^4}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^4+20 d^3 e x-27 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^4-60 d^3 e x+64 d^2 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^4+60 d^3 e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {(4 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^3} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}-\frac {(2 e) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{d^3} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^3 e} \\ & = -\frac {8 e (d-e x)}{5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 e (5 d-8 e x)}{15 d^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (60 d-79 e x)}{15 d^4 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^4 x}+\frac {4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=-\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (15 d^3+149 d^2 e x+222 d e^2 x^2+94 e^3 x^3\right )}{x (d+e x)^3}-60 \sqrt {d^2} e \log (x)+60 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{15 d^5} \]

[In]

Integrate[Sqrt[d^2 - e^2*x^2]/(x^2*(d + e*x)^4),x]

[Out]

-1/15*((d*Sqrt[d^2 - e^2*x^2]*(15*d^3 + 149*d^2*e*x + 222*d*e^2*x^2 + 94*e^3*x^3))/(x*(d + e*x)^3) - 60*Sqrt[d
^2]*e*Log[x] + 60*Sqrt[d^2]*e*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/d^5

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.39

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{4} x}+\frac {4 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{3} \sqrt {d^{2}}}-\frac {19 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e \,d^{3} \left (x +\frac {d}{e}\right )^{2}}-\frac {79 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 d^{4} \left (x +\frac {d}{e}\right )}-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{2} d^{2} \left (x +\frac {d}{e}\right )^{3}}\)	\(199\)
default	\(\frac {-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{d^{2} x}-\frac {2 e^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{d^{2}}}{d^{4}}-\frac {4 e \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )}{d^{5}}+\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}}{e^{2} d^{2}}+\frac {-\frac {3 \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {3 e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}}{d^{4}}+\frac {4 e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d^{5}}-\frac {2 \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{2} d^{4} \left (x +\frac {d}{e}\right )^{3}}\)	\(499\)

[In]

int((-e^2*x^2+d^2)^(1/2)/x^2/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-(-e^2*x^2+d^2)^(1/2)/d^4/x+4*e/d^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-19/15/e/d^3/(
x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-79/15/d^4/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-2/5/e^2/d
^2/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=-\frac {104 \, e^{4} x^{4} + 312 \, d e^{3} x^{3} + 312 \, d^{2} e^{2} x^{2} + 104 \, d^{3} e x + 60 \, {\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (94 \, e^{3} x^{3} + 222 \, d e^{2} x^{2} + 149 \, d^{2} e x + 15 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{3} x^{4} + 3 \, d^{5} e^{2} x^{3} + 3 \, d^{6} e x^{2} + d^{7} x\right )}} \]

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/15*(104*e^4*x^4 + 312*d*e^3*x^3 + 312*d^2*e^2*x^2 + 104*d^3*e*x + 60*(e^4*x^4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2
+ d^3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (94*e^3*x^3 + 222*d*e^2*x^2 + 149*d^2*e*x + 15*d^3)*sqrt(-e^2
*x^2 + d^2))/(d^4*e^3*x^4 + 3*d^5*e^2*x^3 + 3*d^6*e*x^2 + d^7*x)

Sympy [F]

\[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=\int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{2} \left (d + e x\right )^{4}}\, dx \]

[In]

integrate((-e**2*x**2+d**2)**(1/2)/x**2/(e*x+d)**4,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**2*(d + e*x)**4), x)

Maxima [F]

\[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=\int { \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )}^{4} x^{2}} \,d x } \]

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)/((e*x + d)^4*x^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (128) = 256\).

Time = 0.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.13 \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=\frac {4 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{4} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, d^{4} x {\left | e \right |}} + \frac {{\left (15 \, e^{2} + \frac {491 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x} + \frac {1690 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{2} x^{2}} + \frac {2570 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{4} x^{3}} + \frac {1815 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{6} x^{4}} + \frac {555 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{8} x^{5}}\right )} e^{2} x}{30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^2/(e*x+d)^4,x, algorithm="giac")

[Out]

4*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^4*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^
2 + d^2)*abs(e))/(d^4*x*abs(e)) + 1/30*(15*e^2 + 491*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/x + 1690*(d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))^2/(e^2*x^2) + 2570*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^4*x^3) + 1815*(d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))^4/(e^6*x^4) + 555*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^8*x^5))*e^2*x/((d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))*d^4*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d^2-e^2 x^2}}{x^2 (d+e x)^4} \, dx=\int \frac {\sqrt {d^2-e^2\,x^2}}{x^2\,{\left (d+e\,x\right )}^4} \,d x \]

[In]

int((d^2 - e^2*x^2)^(1/2)/(x^2*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(1/2)/(x^2*(d + e*x)^4), x)